计算当前值与一年前值之间的百分比变化 R

Question

这是我的数据框示例：

A = data.frame(retailer = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2),
store = c(5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6), 
week = c(2021100301, 2021092601, 2021091901, 2021091201, 2021082901, 2021082201, 2021081501, 2021080801,
          2020100101, 2020092501, 2020091801, 2020091101, 2020090401, 2020082701, 2020082001, 2020081301),
dollars = c(121817.9, 367566.7, 507674.5, 421257.8, 453330.3, 607551.4, 462674.8,
  464329.1, 339342.3, 549271.5, 496720.1, 554858.7, 382675.5,
  373210.9, 422534.2, 381668.6),
final_week = c("20211040", "20211039", "20211038", "20211037", "20210935", "20210934", "20210933", "20210832", 
          "20201040", "20201039", "20201038", "20201037", "20200935", "20200934", "20200933", "20200832"),
fill = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1))

我先按零售商、然后按商店、然后按每个组中的 final_week 值降序对这些进行分组。如何在每个商店分组中找到最近 4 个 final_week 值与其对应的一年前 final_week 值之间的美元百分比差异？例如，final_week 列中的第一个值是 20211040，它指的是 2021 年。一年前的值是 20201040（2020 年）。我想找到这些值之间的百分比差异，对最近的四个 final_week 值（每个商店组的前四个，因为我按降序排列它们）。

我已开始执行以下操作：

A = A %>%
group_by(retailer, store) %>%
arrange(arrange(retailer, store, desc(final_week), by_group = TRUE) %>%

但我不确定从这里去哪里。谢谢！

Answer 1

我稍微修改了您的示例数据以获得一些有用的结果（一个商店，两年），这里是tidyverse 解决方案：

library(dyplr)
library(tidyr)

A %>%
  # add week and year
  mutate(year_ = substr(final_week, 1,4),
         week_ = substr(final_week,5,8)) %>%
  # remove useless columns
  select(-final_week,- week) %>%
  # from long to wide, with columns for years
  pivot_wider(names_from = year_,
              values_from = c(dollars)) %>%
   # get the last 4 weeks both years
   top_n(wt = week_,4) %>%
   # percentage
   mutate(perc = (`2021`- `2020`)/ `2020`*100)

# A tibble: 4 x 7
  retailer store  fill week_  `2021`  `2020`   perc
     <dbl> <dbl> <dbl> <chr>   <dbl>   <dbl>  <dbl>
1        2     5     1 1040  121818. 339342. -64.1 
2        2     5     1 1039  367567. 549272. -33.1 
3        2     5     1 1038  507674. 496720.   2.21
4        2     5     1 1037  421258. 554859. -24.1 

编辑：

这里有一个多年的解决方案，可能会有所不同：

A %>%
  mutate(year_ = ifelse(substr(final_week,1,4) == min(substr(final_week,1,4)), 'first_y', 'second_y'),
         week_ = substr(final_week,5,8)) %>%
  select(-final_week,- week) %>%
  pivot_wider(names_from = year_,
              values_from = c(dollars)) %>%
   top_n(wt = week_,4) %>%
   mutate(perc = (second_y - first_y)/ first_y*100)

  # A tibble: 4 x 7
  retailer store  fill week_ second_y first_y   perc
     <dbl> <dbl> <dbl> <chr>    <dbl>   <dbl>  <dbl>
1        2     5     1 1040   121818. 339342. -64.1 
2        2     5     1 1039   367567. 549272. -33.1 
3        2     5     1 1038   507674. 496720.   2.21
4        2     5     1 1037   421258. 554859. -24.1 

---

有数据

A <- data.frame(retailer = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2),
               store = rep(5,16), 
               week = c(2021100301, 2021092601, 2021091901, 2021091201, 2021082901, 2021082201, 2021081501, 2021080801,
                        2020100101, 2020092501, 2020091801, 2020091101, 2020090401, 2020082701, 2020082001, 2020081301),
               dollars = c(121817.9, 367566.7, 507674.5, 421257.8, 453330.3, 607551.4, 462674.8,
                           464329.1, 339342.3, 549271.5, 496720.1, 554858.7, 382675.5,
                           373210.9, 422534.2, 381668.6),
               final_week = c("20211040", "20211039", "20211038", "20211037", "20210935", "20210934", "20210933", "20210832", 
                              "20201040", "20201039", "20201038", "20201037", "20200935", "20200934", "20200933", "20200832"),
               fill = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1,1,1,1,1,1,1))